Question

Let $f\left(x\right)$ b ea function defined on $R$ such that $f^{\prime }\left(x\right)=f^{\prime }(3-x),\forall x\in [0,3]$ with $f\left(0\right)=-32$ and $f\left(3\right)=46$, then find the value of ${\int }_0^{3}f\left(x\right)dx$=

• A. $14$
• B. $42$
• C. $21$
• D. $0$

Answer 1

The correct option is C $21$

$\text{Solve: Given,}$

$f^{\prime }\left(x\right)=f^{\prime }(3-x)\forall x\in [0,3]$

$f\left(0\right)=-32andf\left(3\right)=46$

$\text{Let,}I={\int }_0^{3}f\left(x\right)={\int }_0^{3}1\cdot f\left(x\right)dx$

$\text{applying by parts we get,}$

$I=\left[f\right(x)\cdot x{]}_0^3-{\int }_0^3{f}^{\prime }(x)\cdot xdx$

$\Rightarrow I=3f\left(3\right)-0.f\left(0\right)-{\int }_0^{3}x{f}^{\prime }\left(x\right)dx$

$\Rightarrow I=3\times 46-0-I^{\prime }-\left(i\right)$

we have take $I^{\prime }={\int }_0^{3}x{f}^{\prime }\left(x\right)dx$

$\Rightarrow I^{\prime }={\int }_0^{3}x\cdot f^{\prime }\left(x\right)dx-\left(ii\right)$

$x\to 3+0-x$

$x\to 3-x$

$\Rightarrow I^{\prime }={\int }_0^3(3-x)\cdot f^{\prime }(3-x)dx$

$\Rightarrow I^{\prime }={\int }_0^3(3-x)f^{\prime }\left(x\right)dx-$ (iii)$

on adding eqn. (ii) and (iii) we get,

$2I^{\prime }=3{\int }_8^3{f}^{\prime }\left(x\right)dx$

$\Rightarrow I^{\prime }=\frac{3}{2}\left[f\right(x){]}_0^3=\frac{3}{2}[f\left(3\right)-f\left(0\right)]$

$\Rightarrow I^{\prime }=\frac{3}{2}[46+32]$

$\Rightarrow I^{\prime }=\frac{3}{2}\times 78=117$

on putting the value of $I$' in equation

(i) we get

$\Rightarrow I=138-117$

$\Rightarrow I=21$